Question

Prove that secA+tanA-1/tanA-secA+1=cosA/1-sinA

Answer 1

Answer:

hi dear..

Step-by-step explanation:

secA+tanA-1)/tanA-secA+1)

We know that sec^2-tan^2=1

=(secA+tanA-(sec^2 A - tan ^2A)) /tanA-secA+1)

We also know that a^2-b^2=(a+b)(a-b)

= (sec A + tan A - (sec A + tan A) ( sec A - tan A)) / ( tanA-secA+1)

= ( sec A + tan A ) ( 1- (sec A - tan A)/ ( tanA-secA+1)

= (sec A + tan A)/(+ tan A - sec A)/(tan A- sec A+ 1)

= (sec A + tan A)

= 1/cos A + sin A/ cos A

= (1+ sin A)/ cos A

= (1 + sin A )(1- sin A)/(cos A (1- sin A))

= (1- sin ^2 A/(cos A (1- sin A))

= cos ^2 A / (cos A (1- sin A))

= cos A /(1- sin A)

Hence proved

hope it helps

have a great day

Answer 2

${\bold{\blue{\underline{\red{An}\pink{sw}\green{er}\purple{:-}}}}}$

$\rm\:LHS=\dfrac{secA+tanA-1}{tanA-secA+1}$

$\rm\:=\dfrac{(SecA+tanA)-{(sec}^{2}A-{tan}^{2}A)}{(tanA-secA+1)}$

$\rm\:=\dfrac{(SecA+tanA)\:1-(SecA-tanA)}{(tanA-secA+1)}$

$\rm\:=\dfrac{(secA+tanA)(tanA-secA+1)}{(tanA-secA+1)}=(secA+tanA)$

$\rm\:=\bigg(\dfrac{1}{CosA}+\dfrac{SinA}{cosA}\bigg)=\dfrac{(1+sinA)}{(cosA)}\times\:\dfrac{(1-sinA)}{(1-sinA)}$

$\rm\:=\dfrac{(1-{sin}^{2}A)}{cosA(1-sinA)}=\dfrac{{cos}^{2}A}{cosA(1-sinA)}=\dfrac{cosA}{(1-sinA)}=RHS$

$\rm\:=\dfrac{secA+tanA-1}{tanA-secA+1}=\dfrac{cosA}{(1-sinA)}$

$\rm\:LHS=RHS$

$\rm\:hence\:proved$

$\rm\boxed\star\pink{\underline{\underline{{More\: Information.....!!!!!!}}}}$

Some trigonometric identities :-

• $\rm\:{sin}^{2}\theta+{cos}^{2}\theta=1$

• $\rm\:1+{tan}^{2}\theta={sec}^{2}\theta$

• $\rm\:1+{cot}^{2}\theta={cosec}^{2}\theta$

• $\rm\:tan\theta=\dfrac{sin\theta}{cos\theta}$

• $\rm\:tan\theta\:\times\:cot\theta=1$